What are the intercepts of #-6y+8x=1#?

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May 24, 2017

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The intercepts are: #(1/8, 0) and (0, -1/6)#

Explanation:

In order to find the #x- and y-#intercepts, evaluate the other variable at #0#

So, if you put #x = 0#, the function becomes: #-6y + 8(0) = 1#,

or just #-6y = 1#. Now divide by #-6# and you will get:

#y = -1/6.# This is the #y#-intercept.

Next, evaluate the function for # y = 0" "rarr -6(0) +8x = 1#

# 8x = 1" "# Divide by #8# to get:

#x = 1/8#. This is the #x#-intercept.

So, the points are: #(1/8, 0) and (0, -1/6).#

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